Mutual force of electrostatic interaction

I don't think this is an "Easy" problem, unless there's some shortcut I'm not aware of. Anyway, here's a "not so easy" derivation of the result.

Wire 1 has linear charge density $\lambda_1$ , and Wire 2 has linear charge density $\lambda_2$ . The wires are separated by a distance $d$ . $x$ is the position along Wire 2, relative to the position of Wire 1. From Gauss's Law, we know that the electric field strength from Wire 1 is:

$\large{E_1 = \frac{\lambda_1}{2 \pi \epsilon_0 r} = \frac{\lambda_1}{2 \pi \epsilon_0 \sqrt{x^2 + d^2}}}$

The horizontal component is (we needn't consider the vertical component):

$\large{E_{1H} = E_1 \frac{d}{\sqrt{x^2 + d^2}} = \frac{\lambda_1 d}{2 \pi \epsilon_0 (x^2 + d^2)}}$

The horizontal force on an infinitesimal segment of Wire 2 is:

$\large{dF_{H} = E_{1H} \, \lambda_2 \, dx = \frac{\lambda_1\, \lambda_2\, d\, dx}{2 \pi \epsilon_0 (x^2 + d^2)}}$

The total horizontal force is:

$\large{\frac{\lambda_1\, \lambda_2\, d }{2 \pi \epsilon_0} \int_{-\infty}^{\infty} \frac{dx}{x^2 + d^2} = \frac{\lambda_1\, \lambda_2\, d }{2 \pi \epsilon_0} \frac{1}{d} tan^{-1} (\frac{x}{d}) \Big |_{-\infty}^{\infty} = \frac{\lambda_1\, \lambda_2 }{2 \pi \epsilon_0} (\frac{\pi}{2} - (-\frac{\pi}{2}) = \frac{\lambda_1\, \lambda_2 }{2 \epsilon_0} }$

( 2x20)/2ε° = F